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- 2076 Asoj · 5 marks
Graphical method: max and min
Short answer
Plot constraints to find the feasible region, evaluate objective Z = 20x + 10y at corner points. Maximum Z = 800 occurs at (40, 0); Minimum Z = 240 at (6, 12).
Find the maximum and minimum values of the function Z = 20x + 10y subject to x + 2y ≤ 40, 3x + y ≥ 30, 4x + 3y ≥ 60, x, y ≥ 0 by the graphical method.
Step 1: Draw the constraint lines. Replace each inequality by an equation, find the intercepts and decide which side of the line is allowed (test the origin).
| Constraint | Line | Meets x-axis | Meets y-axis | Allowed side |
|---|---|---|---|---|
| (1) x + 2y ≤ 40 | x + 2y = 40 | (40, 0) | (0, 20) | towards the origin |
| (2) 3x + y ≥ 30 | 3x + y = 30 | (10, 0) | (0, 30) | away from the origin |
| (3) 4x + 3y ≥ 60 | 4x + 3y = 60 | (15, 0) | (0, 20) | away from the origin |
Also x ≥ 0 and y ≥ 0, so the region lies in the first quadrant.
Step 2: Feasible region. The region satisfying all constraints is the shaded polygon below (dots), with corner points marked A, B, C, ….
21|
|
| D
| .
| ....
| ......
| ........
| A.........
| ............
| .............
| ...............
| ................
| ..................
| ...................
| .....................
| ......................
0| B.......................C
+--------------------------------------------
0 45
Step 3: Corner points and the objective value. The optimum of a linear function over a convex polygon occurs at a corner point.
| Point | Coordinates | Lines meeting there | Z = 20x + 10y |
|---|---|---|---|
| A | (6, 12) | (2), (3) | 240 |
| B | (15, 0) | (3), y = 0 | 300 |
| C | (40, 0) | (1), y = 0 | 800 |
| D | (4, 18) | (1), (2) | 260 |
Step 4: Conclusion. The largest value is 800 at C(40, 0) and the smallest value is 240 at A(6, 12).
Answer: Maximum Z = 800 at x = 40, y = 0; Minimum Z = 240 at x = 6, y = 12.
✓ Answer checked by Claude Opus 5.5 (high effort)
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